Subject: Chemistry — Ionic Equilibrium
Question: Concentration of $H_{2}SO_{4}$ and $N_{2}SO_{4}$ in a
solution is $1\ M$ and $1.8 \times 10^{- 2}\ M,$ respectively. Molar
solubility of $PbSO_{4}$ in the same solution is $X \times 10^{- Y}\ M$
(expressed in scientific notation). The value of Y is \_\_\_\_\_\_\_.
\[Given: Solubility product of
$PbSO_{4}\left( K_{sp} \right) = 1.6 \times 10^{- 8}$. For
$H_{2}SO_{4},K_{a_{1}}$ is very large and
$K_{a_{2}} = 1.2 \times 10^{- 2}$\]
Answer: (6)
Solution: $\left\lbrack H_{2}SO_{4} \right\rbrack = 1M$ &
$\left\lbrack Na_{2}S_{4} \right\rbrack = 1.8 \times 10^{- 2}\ M$
$$H_{2}SO_{4} \rightarrow H^{+} + HSO_{4}^{1 -}$$
1M 1 M 1 M
$H^{+} + SO_{4}^{2 -}$ $\rightarrow$ $HSO_{4}^{1 -}$
$K = \frac{1}{K_{a_{2}}} = \frac{10^{2}}{1.2}$
1M $1.8 \times 10^{- 2}$ 1M
(1-x) $\left( 1.8 \times 10^{- 2} - x \right)$ (1+x)
$$K = \frac{(1 + x)}{(1 - x)\left( 1.8 \times 10^{- 2} - x \right)} = \frac{100}{1.2}$$
⇒ $1.2 + 1.2x = (100 - 100x)\left( 1.8 \times 10^{- 2}x \right)$
$1.2 + 1.2x = 1.8 - 1.8x - 100x + 100x^{2}$
$100x^{2} - 100.6x + 0.6 = 0$
$x = \frac{100.6 \pm \sqrt{(100.6)^{2} - 4 \times 100 \times 0.6}}{2 \times 100}$
$= \frac{100.6 \pm \sqrt{10120.36 - 240}}{200} = \frac{100.6 \pm 99.4}{200}$
$x = 0.006 = 0.6 \times 10^{- 2}$
$\left\lbrack SO_{4}^{2 -} \right\rbrack = \left( 1.08 \times 10^{- 2} - 0.6 \times 10^{- 2} \right) = 1.2 \times 10^{- 2}$
$K_{sp}\left( PbSO_{4} \right) = \left\lbrack Pb^{2 +} \right\rbrack\left\lbrack SO_{4}^{2 -} \right\rbrack = 1.6 \times 10^{- 8}$
$= \left\lbrack Pb^{2 +} \right\rbrack \times 1.2 \times 10^{- 2} = 1.6 \times 10^{- 8}$
$\left\lbrack Pb^{2 +} \right\rbrack = \frac{1.6}{1.2} \times 10^{- 6} = \frac{4}{3} \times 10^{- 6}$
$$Y = 6$$